$f(x) = 3x-3$ $g(t) = -t^{2}-2t-3(f(t))$ $h(x) = 2x-3(g(x))$ $ f(h(1)) = {?} $
First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = (2)(1)-3(g(1))$ To solve for the value of $h$ , we need to solve for the value of $g(1)$ $g(1) = -1^{2}+(-2)(1)-3(f(1))$ To solve for the value of $g$ , we need to solve for the value of $f(1)$ $f(1) = (3)(1)-3$ $f(1) = 0$ That means $g(1) = -1^{2}+(-2)(1)+(-3)(0)$ $g(1) = -3$ That means $h(1) = (2)(1)+(-3)(-3)$ $h(1) = 11$ Now we know that $h(1) = 11$ . Let's solve for $f(h(1))$ , which is $f(11)$ $f(11) = (3)(11)-3$ $f(11) = 30$